public class SimpleLinkedList<E> {
public Node<E> head;
public int size;
public void add(E e) {
++this.size;
if (null == head) {
this.head = new Node();
head.val = e;
} else {
Node<E> newNode = new Node();
newNode.val = e;
newNode.next = head;
this.head = newNode;
}
}
public void swap(E val1, E val2) {
if (val1.equals(val2)) {
return;
}
Node prevX = null, curr1 = head;
while (curr1 != null && !curr1.val.equals(val1)) {
prevX = curr1;
curr1 = curr1.next;
}
Node prevY = null, curr2 = head;
while (curr2 != null && !curr2.val.equals(val2)) {
prevY = curr2;
curr2 = curr2.next;
}
if (curr1 == null || curr2 == null) {
return;
}
if (prevX == null) {
head = curr2;
} else {
prevX.next = curr2;
}
if (prevY == null) {
head = curr1;
} else {
prevY.next = curr1;
}
Node temp = curr1.next;
curr1.next = curr2.next;
curr2.next = temp;
}
public void reverse() {
Node<E> prev = null;
Node<E> current = head;
Node<E> next = null;
while (current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
head = prev;
}
public static class Node<E> {
public Node<E> next;
public E val;
}
}
public class SimpleLinkedListTest {
#Test
public void testReverseMethod() {
SimpleLinkedList<Integer> myList = new SimpleLinkedList<>();
for (int i = 0; i < 10; i++) {
myList.add(i);
}
SimpleLinkedList<Integer> expectedList = new SimpleLinkedList<>();
for (int i = 9; i > -1; i--) {
expectedList.add(i);
}
myList.reverse();
assertTrue(AssertCustom.assertSLLEquals(expectedList, myList));
}
}
What would be the most optimal way to reverse generic LinkedList by using the swap method?
before reverse method :
(head=[9])->[8]->[7]->[6]->[5]->[4]->[3]->[2]->[1]->[0]-> null
after reverse() method :
(head=[0])->[1]->[2]->[3]->[4]->[5]->[6]->[7]->[8]->[9]-> null
What you need to do is divide the list in half. If the list size is odd the one in the middle will remain in place. Then swap elements on either side in a mirror like fashion. This should be more efficient than O(n^2)
reverse(){
Node current = this.head;
int half = this.size/2;
int midElement = this.size % 2 == 0 ? 0: half + 1;
Stack<Node<E>> stack = new Stack<Node<E>>();
for(int i = 0; i < this.size; i++){
if (i < = half)
stack.push(current);
else{
if (i == midElement)
continue;
else
swap(stack.pop(), current);
current = current.next;
}
}
swap(Node<E> v, Node<E> v1){
E tmp = v.value;
v.value = v1.value;
v1.value = tmp;
}
This is a little bit of pseudo java. It is missing still the checks for size = 0 or size = 1 when it should return immediately. One for loop. Time Complexity O(n). There is also the need to check when size = 2, swap(...) is to be invoked directly.
Based on the #efekctive 's idea, there a solution. The complexity is a little bit worse than O^2 but no need changes in the swap method, no need in usage of another collection. The code below passes the unit test, however, be careful to use it there could be a bug related to size/2 operation. Hope this help.
public void reverse() {
Node<E> current = head;
SimpleLinkedList<E> firstHalf = new SimpleLinkedList<>();
SimpleLinkedList<E> secondHalf = new SimpleLinkedList<>();
for (int i = 0; i < size; i++) {
if (i >= size / 2) {
firstHalf.add(current.val);
} else {
secondHalf.add(current.val);
}
current = current.next;
}
SimpleLinkedList<E> secondHalfReverse = new SimpleLinkedList<>();
for (int i = 0; i < secondHalf.size(); i++) {
secondHalfReverse.add(secondHalf.get(i));
}
for (int i = 0; i < size / 2; i++) {
if (secondHalfReverse.get(i) == firstHalf.get(i)) {
break;
}
swap(secondHalfReverse.get(i), firstHalf.get(i));
}
}
Related
In this linked list I am trying to delete a node and return the list after deleting the particular node. Here I am not using the Value to delete, I am using the position to delete that particular node. Here in my code the delete function seems to have no effect over the output. What am I doing wrong here?
import java.util.*;
class LinkedList{
Node head;
static class Node{
int data;
Node next;
Node(int d){
this.data = d;
next = null;
}
}
public LinkedList insert(LinkedList l, int data){
Node new_node = new Node(data);
if(l.head == null){
l.head = new_node;
}
else{
Node last = l.head;
while(last.next != null){
last = last.next;
}
last.next = new_node;
}
return l;
}
public LinkedList delete(LinkedList l, int position){
Node current = l.head;
if(position == 0){
current = current.next;
}
int index = 1;
while(index < position - 1){
current = current.next;
index++;
}
current = current.next.next;
Node iterating = l.head;
while(iterating != null){
System.out.print(iterating.data + " ");
iterating = iterating.next;
}
return l;
}
}
public class Main
{
public static void main(String[] args) {
LinkedList l = new LinkedList();
Scanner sc = new Scanner(System.in);
int number = sc.nextInt();
int position = sc.nextInt();
for(int i=0; i<number; i++){
int num = sc.nextInt();
l.insert(l,num);
}
l.delete(l,position);
}
}
current=current.next doesn't have any effect on the original LinkedList because with that line of code, you just change where the reference (named as current) points to.
Think about this way,
Node a = new Node()
Node b = new Node()
Node c = new Node()
a.next =b
a = c
These lines of code doesn't result in c being connected to b.
Change code to this:
if (position == 0) return l.head.next;
else {
Node head = l.head;
int index = 1;
Node itr = l.head;
while(index < position) {
itr = itr.next;
index++;
}
if(itr.next != null) itr.next = itr.next.next;
return head;
}
public LinkedList delete(LinkedList l, int position) {
Node previous = null;
Node current = l.head;
int index = 0;
while (current != null && index < position){
previous = current;
current = current.next;
index++;
}
if (current != null) {
if (previous == null) {
l.head = current.next;
} else {
previous.next = current.next;
}
}
System.out.print("[");
Node iterating = l.head;
while (iterating != null) {
System.out.print(iterating.data + ", ");
iterating = iterating.next;
}
System.out.println("]");
return l;
}
The problem in java is that to delete a node either the head must be changed to its next, or the previous node's next must be changed to current's next.
Then too current might become null, have reached the list's end, position > list length.
My problem occurs when I track lastC, the last node in my linked list. For some reason, outside of the method it seems to be null constantly. I want to see if there is a better way to track it to improve run time on my program.
Append:
public MyStringBuilder append(String s)
{
if(this.length == 0) {
firstC = new CNode(s.charAt(0));
length = 1;
CNode currNode = firstC;
// Create remaining nodes, copying from String. Note
// how each new node is simply added to the end of the
// previous one. Trace this to see what is going on.
for (int i = 1; i < s.length(); i++)
{
CNode newNode = new CNode(s.charAt(i));
currNode.next = newNode;
currNode = newNode;
length++;
}
lastC = currNode;
//System.out.println(lastC.data);
//lastC.next = null;//new
} else {
if(lastC == null) {
lastC = getCNodeAt(length - 1);
//System.out.println("no");
}
//System.out.println(lastC.data);
//CNode currNode = lastC;
//CNode currNode = firstC;//changed
//for(int j = 0; j < length - 1; j++) {
// currNode = currNode.next;
//}
CNode currNode = lastC;
//lastC = currNode;
for (int i = 0; i < s.length(); i++)
{
CNode newNode = new CNode(s.charAt(i));
currNode.next = newNode;
currNode = newNode;//
length++;
}
lastC = currNode;
}
//lastC = getCNodeAt(length - 1);
return this;
}
Constructor:
public MyStringBuilder(String s)
{
if (s == null || s.length() == 0) // Special case for empty String
{ // or null reference
firstC = null;
lastC = null;
length = 0;
}
else
{
// Create front node to get started
firstC = new CNode(s.charAt(0));
length = 1;
CNode currNode = firstC;
// Create remaining nodes, copying from String. Note
// how each new node is simply added to the end of the
// previous one. Trace this to see what is going on.
for (int i = 1; i < s.length(); i++)
{
CNode newNode = new CNode(s.charAt(i));
currNode.next = newNode;
currNode = newNode;
length++;
}
lastC = currNode;
//lastC = getCNodeAt(length - 1);
//System.out.println(lastC.data);
}
}
Instance variables
private CNode firstC; // reference to front of list. This reference is necessary
// to keep track of the list
private CNode lastC; // reference to last node of list. This reference is
// necessary to improve the efficiency of the append()
// method
private int length;
Suggestions:
use sentinel node to get rid of unnecessary ifs
conventionally, append() should be void
constructor calls append() since they are doing similar jobs
Not a Java programmer, but something like this should work:
public class MyStringBuilder {
private CNode sentinel = new CNode('s');
private CNode lastC = sentinel;
private int length = 0;
public MyStringBuilder(String s){
append(s);
}
public void append(String s){
for (int i = 1; i < s.length(); i++){
CNode newNode = new CNode(s.charAt(i));
lastC.next = newNode;
lastC = newNode;
length++;
}
}
public CNode getHead(){
return sentinel.next;
}
public CNode getTail(){
if (length) {
return lastC;
} else {
return null;
}
}
}
I am needing to sort a linked list alphabetically. I have a Linked List full of passengers names and need the passengers name to be sorted alphabetically. How would one do this? Anyone have any references or videos?
You can use Collections#sort to sort things alphabetically.
In order to sort Strings alphabetically you will need to use a Collator, like:
LinkedList<String> list = new LinkedList<String>();
list.add("abc");
list.add("Bcd");
list.add("aAb");
Collections.sort(list, new Comparator<String>() {
#Override
public int compare(String o1, String o2) {
return Collator.getInstance().compare(o1, o2);
}
});
Because if you just call Collections.sort(list) you will have trouble with strings that contain uppercase characters.
For instance in the code I pasted, after the sorting the list will be: [aAb, abc, Bcd] but if you just call Collections.sort(list); you will get: [Bcd, aAb, abc]
Note: When using a Collator you can specify the locale Collator.getInstance(Locale.ENGLISH) this is usually pretty handy.
In java8 you no longer need to use Collections.sort method as LinkedList inherits the method sort from java.util.List, so adapting Fido's answer to Java8:
LinkedList<String>list = new LinkedList<String>();
list.add("abc");
list.add("Bcd");
list.add("aAb");
list.sort( new Comparator<String>(){
#Override
public int compare(String o1,String o2){
return Collator.getInstance().compare(o1,o2);
}
});
References:
http://docs.oracle.com/javase/8/docs/api/java/util/LinkedList.html
http://docs.oracle.com/javase/7/docs/api/java/util/List.html
Elegant solution since JAVA 8:
LinkedList<String>list = new LinkedList<String>();
list.add("abc");
list.add("Bcd");
list.add("aAb");
list.sort(String::compareToIgnoreCase);
Another option would be using lambda expressions:
list.sort((o1, o2) -> o1.compareToIgnoreCase(o2));
Here is the example to sort implemented linked list in java without using any standard java libraries.
package SelFrDemo;
class NodeeSort {
Object value;
NodeeSort next;
NodeeSort(Object val) {
value = val;
next = null;
}
public Object getValue() {
return value;
}
public void setValue(Object value) {
this.value = value;
}
public NodeeSort getNext() {
return next;
}
public void setNext(NodeeSort next) {
this.next = next;
}
}
public class SortLinkList {
NodeeSort head;
int size = 0;
NodeeSort add(Object val) {
// TODO Auto-generated method stub
if (head == null) {
NodeeSort nodee = new NodeeSort(val);
head = nodee;
size++;
return head;
}
NodeeSort temp = head;
while (temp.next != null) {
temp = temp.next;
}
NodeeSort newNode = new NodeeSort(val);
temp.setNext(newNode);
newNode.setNext(null);
size++;
return head;
}
NodeeSort sort(NodeeSort nodeSort) {
for (int i = size - 1; i >= 1; i--) {
NodeeSort finalval = nodeSort;
NodeeSort tempNode = nodeSort;
for (int j = 0; j < i; j++) {
int val1 = (int) nodeSort.value;
NodeeSort nextnode = nodeSort.next;
int val2 = (int) nextnode.value;
if (val1 > val2) {
if (nodeSort.next.next != null) {
NodeeSort CurrentNext = nodeSort.next.next;
nextnode.next = nodeSort;
nextnode.next.next = CurrentNext;
if (j == 0) {
finalval = nextnode;
} else
nodeSort = nextnode;
for (int l = 1; l < j; l++) {
tempNode = tempNode.next;
}
if (j != 0) {
tempNode.next = nextnode;
nodeSort = tempNode;
}
} else if (nodeSort.next.next == null) {
nextnode.next = nodeSort;
nextnode.next.next = null;
for (int l = 1; l < j; l++) {
tempNode = tempNode.next;
}
tempNode.next = nextnode;
nextnode = tempNode;
nodeSort = tempNode;
}
} else
nodeSort = tempNode;
nodeSort = finalval;
tempNode = nodeSort;
for (int k = 0; k <= j && j < i - 1; k++) {
nodeSort = nodeSort.next;
}
}
}
return nodeSort;
}
public static void main(String[] args) {
SortLinkList objsort = new SortLinkList();
NodeeSort nl1 = objsort.add(9);
NodeeSort nl2 = objsort.add(71);
NodeeSort nl3 = objsort.add(6);
NodeeSort nl4 = objsort.add(81);
NodeeSort nl5 = objsort.add(2);
NodeeSort NodeSort = nl5;
NodeeSort finalsort = objsort.sort(NodeSort);
while (finalsort != null) {
System.out.println(finalsort.getValue());
finalsort = finalsort.getNext();
}
}
}
Node mergeSort(Node head) {
if(head == null || head.next == null) {
return head;
}
Node middle = middleElement(head);
Node nextofMiddle = middle.next;
middle.next = null;
Node left = mergeSort(head);
Node right = mergeSort(nextofMiddle);
Node sortdList = merge(left, right);
return sortdList;
}
Node merge(Node left, Node right) {
if(left == null) {
return right;
}
if(right == null) {
return left;
}
Node temp = null;
if(left.data < right.data) {
temp = left;
temp.next = merge(left.next, right);
} else {
temp = right;
temp.next = merge(left, right.next);
}
return temp;
}
Node middleElement(Node head) {
Node slow = head;
Node fast = head;
while (fast != null && fast.next != null && fast.next.next != null) {
fast = fast.next.next;
slow = slow.next;
}
return slow;
}
I wouldn't. I would use an ArrayList or a sorted collection with a Comparator. Sorting a LinkedList is about the most inefficient procedure I can think of.
You can do it by Java 8 lambda expression :
LinkedList<String> list=new LinkedList<String>();
list.add("bgh");
list.add("asd");
list.add("new");
//lambda expression
list.sort((a,b)->a.compareTo(b));
If you'd like to know how to sort a linked list without using standard Java libraries, I'd suggest looking at different algorithms yourself. Examples here show how to implement an insertion sort, another StackOverflow post shows a merge sort, and ehow even gives some examples on how to create a custom compare function in case you want to further customize your sort.
I'm having some trouble with NPE's in a MyLinkedList class that extends AbstractList. I start with these constructors:
The constructor for the private Node class:
public Node(T nodeData, Node<T> nodePrev, Node<T> nodeNext)
{
this.data = nodeData;
this.prev = nodePrev;
this.next = nodeNext;
}
The constructor for the MyLinkedList class
MyLinkedList()
{
this.head = new Node<T>(null, null, null);
this.tail = new Node<T>(null, null, null);
this.size = 0;
}
MyLinkedList(Node<T> head, Node<T> tail, int size)
{
this.head = head;
this.tail = tail;
this.size = size;
}
and here I try to return the node at an index with this method:
private Node<T> getNth(int index)
{
Node<T> temp;
if(index < 0 || index > size)
throw new IndexOutOfBoundsException();
if(index < this.size() / 2)
{
temp = this.head;
for(int i = 0; i < index; i++)
{
temp = temp.getNext();
}
}
else
{
temp = this.tail;
for(int i = this.size(); i > index; i--)
{
temp = temp.getPrev();
}
}
return temp;
}
I think the main problem has something to do with initializing the head and tail as null, but I'm not sure if this is the problem and if it is, how to fix it. Is there a better way to initialize these Nodes to avoid NPE's?
You're initializing both the head and the tail of the list with this:
MyLinkedList()
{
this.head = new Node<T>(null, null, null);
this.tail = new Node<T>(null, null, null);
this.size = 0;
}
This seems to be the main source of your NPE's, because your iteration doesn't do any kind of checks. In particular, your method will fail in border conditions (since you already check for lengths before even trying to iterate).
By adding some checks you can avoid those exceptions:
private Node<T> getNth(int index)
{
Node<T> temp = null; //Always try to initialize your variables if you're going
//to return them.
if(index < 0 || index > size)
throw new IndexOutOfBoundsException();
if(index < this.size() / 2)
{
temp = this.head;
for(int i = 0; i < index; i++)
{
if(temp.getNext() != null)
temp = temp.getNext();
else
break;//Break the iteration if there is not a next node
}
}
else
{
temp = this.tail;
for(int i = this.size(); i > index; i--)
{
if(temp.getPrev() != null)
temp = temp.getPrev();
else
break;
}
}
return temp;
}
You can throw some kind of exception instead of breaking the iterations if you want.
I am needing to sort a linked list alphabetically. I have a Linked List full of passengers names and need the passengers name to be sorted alphabetically. How would one do this? Anyone have any references or videos?
You can use Collections#sort to sort things alphabetically.
In order to sort Strings alphabetically you will need to use a Collator, like:
LinkedList<String> list = new LinkedList<String>();
list.add("abc");
list.add("Bcd");
list.add("aAb");
Collections.sort(list, new Comparator<String>() {
#Override
public int compare(String o1, String o2) {
return Collator.getInstance().compare(o1, o2);
}
});
Because if you just call Collections.sort(list) you will have trouble with strings that contain uppercase characters.
For instance in the code I pasted, after the sorting the list will be: [aAb, abc, Bcd] but if you just call Collections.sort(list); you will get: [Bcd, aAb, abc]
Note: When using a Collator you can specify the locale Collator.getInstance(Locale.ENGLISH) this is usually pretty handy.
In java8 you no longer need to use Collections.sort method as LinkedList inherits the method sort from java.util.List, so adapting Fido's answer to Java8:
LinkedList<String>list = new LinkedList<String>();
list.add("abc");
list.add("Bcd");
list.add("aAb");
list.sort( new Comparator<String>(){
#Override
public int compare(String o1,String o2){
return Collator.getInstance().compare(o1,o2);
}
});
References:
http://docs.oracle.com/javase/8/docs/api/java/util/LinkedList.html
http://docs.oracle.com/javase/7/docs/api/java/util/List.html
Elegant solution since JAVA 8:
LinkedList<String>list = new LinkedList<String>();
list.add("abc");
list.add("Bcd");
list.add("aAb");
list.sort(String::compareToIgnoreCase);
Another option would be using lambda expressions:
list.sort((o1, o2) -> o1.compareToIgnoreCase(o2));
Here is the example to sort implemented linked list in java without using any standard java libraries.
package SelFrDemo;
class NodeeSort {
Object value;
NodeeSort next;
NodeeSort(Object val) {
value = val;
next = null;
}
public Object getValue() {
return value;
}
public void setValue(Object value) {
this.value = value;
}
public NodeeSort getNext() {
return next;
}
public void setNext(NodeeSort next) {
this.next = next;
}
}
public class SortLinkList {
NodeeSort head;
int size = 0;
NodeeSort add(Object val) {
// TODO Auto-generated method stub
if (head == null) {
NodeeSort nodee = new NodeeSort(val);
head = nodee;
size++;
return head;
}
NodeeSort temp = head;
while (temp.next != null) {
temp = temp.next;
}
NodeeSort newNode = new NodeeSort(val);
temp.setNext(newNode);
newNode.setNext(null);
size++;
return head;
}
NodeeSort sort(NodeeSort nodeSort) {
for (int i = size - 1; i >= 1; i--) {
NodeeSort finalval = nodeSort;
NodeeSort tempNode = nodeSort;
for (int j = 0; j < i; j++) {
int val1 = (int) nodeSort.value;
NodeeSort nextnode = nodeSort.next;
int val2 = (int) nextnode.value;
if (val1 > val2) {
if (nodeSort.next.next != null) {
NodeeSort CurrentNext = nodeSort.next.next;
nextnode.next = nodeSort;
nextnode.next.next = CurrentNext;
if (j == 0) {
finalval = nextnode;
} else
nodeSort = nextnode;
for (int l = 1; l < j; l++) {
tempNode = tempNode.next;
}
if (j != 0) {
tempNode.next = nextnode;
nodeSort = tempNode;
}
} else if (nodeSort.next.next == null) {
nextnode.next = nodeSort;
nextnode.next.next = null;
for (int l = 1; l < j; l++) {
tempNode = tempNode.next;
}
tempNode.next = nextnode;
nextnode = tempNode;
nodeSort = tempNode;
}
} else
nodeSort = tempNode;
nodeSort = finalval;
tempNode = nodeSort;
for (int k = 0; k <= j && j < i - 1; k++) {
nodeSort = nodeSort.next;
}
}
}
return nodeSort;
}
public static void main(String[] args) {
SortLinkList objsort = new SortLinkList();
NodeeSort nl1 = objsort.add(9);
NodeeSort nl2 = objsort.add(71);
NodeeSort nl3 = objsort.add(6);
NodeeSort nl4 = objsort.add(81);
NodeeSort nl5 = objsort.add(2);
NodeeSort NodeSort = nl5;
NodeeSort finalsort = objsort.sort(NodeSort);
while (finalsort != null) {
System.out.println(finalsort.getValue());
finalsort = finalsort.getNext();
}
}
}
Node mergeSort(Node head) {
if(head == null || head.next == null) {
return head;
}
Node middle = middleElement(head);
Node nextofMiddle = middle.next;
middle.next = null;
Node left = mergeSort(head);
Node right = mergeSort(nextofMiddle);
Node sortdList = merge(left, right);
return sortdList;
}
Node merge(Node left, Node right) {
if(left == null) {
return right;
}
if(right == null) {
return left;
}
Node temp = null;
if(left.data < right.data) {
temp = left;
temp.next = merge(left.next, right);
} else {
temp = right;
temp.next = merge(left, right.next);
}
return temp;
}
Node middleElement(Node head) {
Node slow = head;
Node fast = head;
while (fast != null && fast.next != null && fast.next.next != null) {
fast = fast.next.next;
slow = slow.next;
}
return slow;
}
I wouldn't. I would use an ArrayList or a sorted collection with a Comparator. Sorting a LinkedList is about the most inefficient procedure I can think of.
You can do it by Java 8 lambda expression :
LinkedList<String> list=new LinkedList<String>();
list.add("bgh");
list.add("asd");
list.add("new");
//lambda expression
list.sort((a,b)->a.compareTo(b));
If you'd like to know how to sort a linked list without using standard Java libraries, I'd suggest looking at different algorithms yourself. Examples here show how to implement an insertion sort, another StackOverflow post shows a merge sort, and ehow even gives some examples on how to create a custom compare function in case you want to further customize your sort.